Which of the following definite integrals are equal to \( \lim _{n \rightarrow \infty} \sum_{k=1}^{n} \sin \left(-1+\frac{5 k}{n}\right) \frac{5}{n} \) ? I. \( \int_{-1}^{4} \sin x d x \) II. \( \int_{0}^{5} \sin (-1+x) d x \) III. \( 5 \int_{0}^{1} \sin (-1+5 x) d x \) (A) I only (B) II only (C) III only (D) I, II, and III

To determine which definite integrals are equal to the limit given, we start by analyzing the Riemann sum. The limit \( \lim_{n \rightarrow \infty} \sum_{k=1}^{n} \sin \left(-1+\frac{5 k}{n}\right) \frac{5}{n} \) can be understood as the definition of the Riemann sum for the function \( \sin(-1 + 5x) \) over the interval from \( 0 \) to \( 1 \). The substitution \( \frac{k}{n} \) generates \( x \) values that range from \( 0 \) to \( 1 \) as \( k \) goes from \( 1 \) to \( n \). Let's rewrite the limit more formally: \[ \lim_{n \to \infty} \sum_{k=1}^{n} \sin\left(-1 + \frac{5k}{n}\right) \cdot \frac{5}{n} = \int_{0}^{1} \sin(-1 + 5x) \, dx \] Now, comparing with the given integrals: I. The integral \( \int_{-1}^{4} \sin x \, dx \) evaluates to the area under the curve of \( \sin x \) from \( -1 \) to \( 4 \). II. The integral \( \int_{0}^{5} \sin(-1 + x) \, dx \) is a transformation of \( \sin \) that can be computed but is over a different interval. III. The integral \( 5 \int_{0}^{1} \sin(-1 + 5x) \, dx \) exactly matches the form we found, considering the limits and factors match up. Thus, the third option directly corresponds to the original limit, while the others do not. The correct answer is (C) III only.